Polygon Problem.

Let $|\overline {\rm AB}| = 2r$ and $n$ be even integer and $n \geq 4$ .

Using the diagram directly above we have:

$x = 2r\sin(\dfrac{\pi}{n})$ and the height $h = r\cos(\dfrac{\pi}{n}) \implies$

The area $A_{n} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) r^2$

Using right $\triangle{ABC}$ in the first diagram above we have:

$(2r)^2 = 21^2 + 3^2 = 450 \implies r = \dfrac{15\sqrt{2}}{2} = \dfrac{15}{\sqrt{2}} \implies$

$A_{n} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) (\dfrac{225}{2}) = \dfrac{225}{4} n\sin(\dfrac{2\pi}{n})$

Using $n = 100 \implies A_{100} =225 * 25 \sin(\dfrac{\pi}{50}) = 5625 \sin(\dfrac{\pi}{50})$

$\approx \boxed{353.196672}$ .

$$

$$

$$

$$

$$

$$

Note:

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1$ we have:

$\pi\cos(\dfrac{2\pi}{n}) < \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) < \pi$ and $\pi\lim_{n \rightarrow \infty} \cos(\dfrac{2\pi}{n}) = \pi$

$\therefore$ by squeeze play theorem $\implies \lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) = \pi$ $\implies$ the area of the circle $A_{c} = \pi r^2$

Now using the value of $r$ above we obtain:

$A_{c} = \dfrac{225}{2}\pi \approx 353.4291735$ .